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受欢迎的微积分 >

integral from 0 to 1 of 2pix(2e^{-x^2})

解答

\int_{0}^{1} 2 π x (2 e^{- x^{2}}) d x

解答

2 π (1 - \frac{1}{e})

+1

十进制

3.97173 \dots

求解步骤

\int_{0}^{1} 2 π x \cdot 2 e^{- x^{2}} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 π 2 \cdot \int_{0}^{1} x e^{- x^{2}} d x

化简

= 4 π \cdot \int_{0}^{1} x e^{- x^{2}} d x

使用换元积分法

= 4 π \cdot \int_{0}^{- 1} - \frac{e ^{u}}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 4 π (- \int_{- 1}^{0} - \frac{e ^{u}}{2} d u)

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 π (- (- \frac{1}{2} \cdot \int_{- 1}^{0} e^{u} d u))

使用积分常见定则:

\int e^{u} d u = e^{u}

= 4 π (- (- \frac{1}{2} [e^{u}]_{- 1}^{0}))

化简

4 π (- (- \frac{1}{2} [e^{u}]_{- 1}^{0})) : 2 π [e^{u}]_{- 1}^{0}

= 2 π [e^{u}]_{- 1}^{0}

带入上下限计算后得:

1 - \frac{1}{e}

= 2 π (1 - \frac{1}{e})

作图

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